4.3: CLASSICAL GENETICS
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1. CLASSICAL GENETICS : DEFINITIONS
Genotype
The alleles of an organism
Gene expression
Homozygous
The transformation of genotype into
phenotype, determined by the allel
ecombination
The characteristics of an organism
(may refer to physical, behavioural,
disease predisposition etc)
Gene will be expressed in the
phenotype when present (CAPITAL
LETTER, e.g. D)
Gene will be ONLY expressed in the
phenotype present in the homozygous
state (SMALL LETTER, e.g. d)
Pairs of alleles that both affect the
phenotype when present in a
heterozygote
Having two identical alleles of a gene
Heterozygous
Having two different alleles of a gene
Carrier
An individual that has one copy of a
recessive allele that causes a genetic
disease in individuals that are
homozygous for this allele
A phenotype created from a
combination of codominant alleles
A chromosome which is not a sex
chromosome (e.g. 1 – 22 in humans)
The chromosomes which determine
gender (e.g. chromosomes X/Y in
humans)
The specific position of a gene on a
chromosome (e.g. 15p.322)
A mating between two individuals who
have potentially different alleles at a
single gene locus
Testing a suspected heterozygote by
crossing it with a known homozygous
recessive
Phenotype
Dominant allele
Recessive allele
Codominant alleles
Mixed phenotype
Autosome
Sex chromosome
Gene locus
Monohybrid cross
Test cross
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2. Outline the conventions for notation of genotypes, giving one example of
each:
Dominant/ Recessive
alleles
D OR d
Codominant alleles
Sex-linked alleles
HbA or HbS
XH or Xh
Huntingdon’s Disease
Cystic fibrosis
Phenylketonuria
Sickle Cell anaemia
Blood type
Sex-linked Haemophilia
A or B
Sex-linked colour
blindness
3. The allele for tall plants (T) is dominant over the allele for dwarf plants (t).
a) State the possible genotypes of a tall plant : TT or Tt
b) Explain how a test cross could be used to determine the genotype of a tall plant:
Test with a known homozygous recessive (tt). If any of the offspring are small
plants, then this proves that the tall plant must be heterozygous.
Mendel is known as the father of genetics for his extensive work with many types
of crosses of pea plants.
Complete the Punnett grid below to show the outcome of the monohybrid cross
between homozygous individuals, that results in peas of different colours
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100% of F1 generation will be Yy (Yellow) [heterozygotes]
d) Now complete a Punnett Square to show the possible outcomes of a cross
between two members of the F1 generation. Describe all genotypes produced.
Genotype of F1 parents: both Yy
Phenotype of parents: both yellow
Genotype of offspring (F2): 25% YY (yellow): 50% Yy (yellow): 25% yy (green)
Phenotype of offspring: 75% yellow: 25% green
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CODOMINANCE I: BLOOD TYPES
Human ABO blood types follow a codominant pattern.
4. Describe what is meant by ‘some genes have multiple alleles’
While there are only 2 alleles of each gene in each diploid cell, there can be more
than two versions of that allele present and thus there are multiple
combinations of alleles (genotypes) possible depending on which allele
combination is present.
5. a) Complete the table below to show how blood type is inherited.
Alleles
i
IA
IB
i
ii
IA i
IB i
IA
IA i
IA IA
IA IB
IB
IB i
IA IA
IB IB
b) Highlight the genotype and phenotype which is an example of codominance
IA IB: Phenotype AB
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PEDIGREE CHARTS
Complete the Pedigree chart to show the inheritance of blood types
Line 1: O is ii: A is IAi
Line 2: A is IAi
Line 3: B is IBi, A is IAi
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CODOMINANCE 2: SICKLE CELL DISEASE
Sickle cell trait/ disease is another example of codominance.
6. State the genotypes and phenotypes of these individuals
Genotype
Phenotype
HbAHbA
Healthy non
carrier
(susceptible to
malaria)
HbAHbs
Sickle cell trait
(carrier)
(malaria resistant)
HbsHbs
Sickle cell disease
(malaria
susceptible)
7. Predict the phenotype ratios of offspring in the following crosses. Show all of
your working, and set it out as expected. Take care with notation.
i)
Carrier mother with affected father
ii)
Affected father with unaffected mother
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iii)
Carrier mother with carrier father
8. Review: Explain how the prevalence of sickle cell anaemia in regions of Africa is
an example of natural selection in action.
Natural selection: selection of offspring with a survival advantage: those that have
the best characteristics for surviving will then breed and thus pass on their genes.
Since HbS confers resistance to malaria, individuals carrying one sickle cell gene
have a survival advantage in areas where there is a lot of malaria; thus they survive
and pass on the gene to the next generation.
Regions where malaria is endemic have a relatively high prevalence of the HbS
allele, since it confers resistance to malaria, and the selection advantage of a single
copy of HbS must outweigh the selection disadvantage of being homozygous for the
trait and developing sickle cell anaemia.
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9. The Pedigree table below shows a family affected by sickle cell disease. Deduce
the genotype of each individual represented by a letter.
A HbAHbS
B HbSHbS
C HbSHbS
D HbAHbS
E HbSHbS
F HbSHbS
G HbAHbS
H HbSHbS
$ HbAHbS or HbAHbA
£ HbAHbS
a) Calculate the probability that any further children produced by E and her partner
≠ will have sickle cell disease.
Partner is a carrier HbAHbS
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b) Male $ is healthy but has unknown genotype. Calculate the probability that any
children produced with D will have sickle cell anaemia. Show all of your working.
$ is EITHER HbAHbS or HbAHbA
D is a carrier for sickle cell disease: HbAHbS
If $ is a carrier:
If $ is homozygous for HbA, HbAHbA, then 50% of offspring will be
HbAHbA and 50% of offspring will be HbAHbS.
Thus the risk of having sickle cell disease is 1/8 = 12.5%
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SEX LINKED TRAITS
Some traits are autosomal while others are carried on sex chromosomes.
10. Distinguish between autosomes and sex chromosomes.
An autosome is any one of the numbered chromosomes, as opposed to the sex
chromosomes (X/Y).
11. Annotate the diagram and distinguish between the X and Y chromosome.
The X chromosome carries approximately 100 genes. Over 100
genes coding for recessive genetic sex-linked diseases have been
identified on the X chromosome, including haemophilia, sex-linked
colour blindness, Duchenne muscular dystrophy.
The Y chromosome carries approximately 50 – 60 protein coding
genes, most of which code for proteins involved in male sex
determination and development. Sexis determined by the SRY gene.
Many genes are unique to the Y chromosome; genes in the so-called
pseudo-autosomal region are found on both X and Y chromosomes.
12. Outline the role of the SRY gene on the Y chromosome.
The SRY gene codes for the SRY protein (gene locus Yp11.3). This is
a protein involved in regulating the formation of the testes, and is
considered one of the principal genes responsible for gender
determination.
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13. Outline how non-disjunction can lead to gender-related chromosomal
abnormalities.
Non disjunction during anaphase I or II of meiosis can lead to abnormalities in
the number of sex chromosomes, most commonly trisomy or monosomy.
XXY (Klinefelters syndrome); 22 XO = monosomy X (Turner’s Syndrome)
Certain genetic disorders are associated with gender.
14. Define sex-linked.
This refers to a disorder where the gene responsible is located on a sex
chromosome (typically the X chromosome). Many but not all sex linked diseases
are recessive traits. Very few diseases are caused be genes located on the Y
chromosome.
15. State two classic examples of sex-linked diseases
Haemophilia A, haemophilia B, sex-linked red-green colour blindness, Duchenne
Muscular dystrophy, severe combined immunideficiancy syndrome (SCIDs).
16. Explain why sex-linked diseases are more common in males than females.
For a recessive condition, males need only have a single copy of the recessive
allele in order to express the genotype/phenotype (have the disease), whereas
females need to have both recessive alleles in order to develop the condition.
Consequently, the % incidence of a recessive sex-linked condition in females is
equal to the square of its incidence in males: for example,if the incidence is 50%
in males, it would be 50 X 50% in females, i.e. 25%....if it were 1 in 20 in males it
would be 1 in 300 in females.
17. Explain why females can be homozygous or heterozygous for a recessive sexlinked trait, whereas males cannot.
Males only have one X chromosome, so are either disease free (carrying the
dominant allele, XHy) or disease positive (carrying the recessive allele, XhY
For a recessive condition, males need only have a single copy of the recessive
allele in order to express the genotype/phenotype (have the disease), whereas
females need to have both recessive alleles in order to develop the condition.
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18. The allele for colour-blindness (n) is recessive to the allele for normal vision (N).
This gene is carried on a non-homologous region on the X chromosome. Complete
the table below to show the genotypes and phenotypes of the offspring with regard
to colour -blindness.
Normal
Female
XNXN
Male
X NY
Affected
Xn Xn
XnY
Carrier
XN Xn
possible or not?NO
19. Complete a Punnett Square to show a cross between a normal male and carrier
female. What is the expected ratio of phenotypes?
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HAEMOPHILIA IS A SEX-LINKED AND RECESSIVE DISORDER.
20. What is the normal function of the gene associated with haemophilia?
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Haemophilia A is caused by a mutation in the gene coding for Factor VIII
which is an essential factor in the blood clotting cascade.
Haemophilia A is caused by a mutation in the gene coding for Factor IX which
is an essential factor in the blood clotting cascade.
21. Describe the effects and symptoms of haemophilia
Haemophilia is abnormal clotting ability of the blood; individuals with haemophilia
cannot clot blood following minor or major injury, and so will develop bruises or
severe haemorrhages easily. Many risk death from internal or brain haemorrhage.
22. Outline one form of genetic engineering used to help individuals with
haemophilia.
Polly and Molly are 2 genetically engineered sheep, which produce Factor IX in
their milk.
23. Suggest reasons why the frequency of some disease-related alleles may be
increasing in the population.
We now have medical treatments that allow individuals with genetic diseases to
survive, and thus reproduce and pass their genes on to future generations. In
the past, these individuals would have succumbed (died) from many of these
diseases, so that the genes would not have been passed on through
reproduction and would have gradually diminished in the population.
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